A ball is thrown vertically upward with an initial velocity of 96 feet per second

a basketball is thrown vertically up with an initial velocity
t = 6 if basketball achieves underneath
A) At what time t will the basketball strike the base?
it
t = 6t=0when the basketball is thrown
t – 6 = 0
pointsthanks the support
=>
distanceof the basketball from floor = 128feet.so, 96t – 16t2 = 128
t – 2 = 0
4 or

afterthrowing.Answer:- after
=>
it
Along basketball through floor after t sec exists bythe
– 4t- 2t + 8 = 0
theground? Step by step on which you performed this recieve life-saver
=> 16t(t – 6) =0henceeither, 16t
A)6
B)2
– 6t+ 8 = 0

Step by step on which you performed this recieve life-saver
A) the distance of basketball through groundafter t sec is
=>
6 moments of throwing the basketball
s(t) = 96t – 16t2at thetime of hitting underneath the distance of basketball
6 moments of throwing the basketball
=0 or
s(t) = 96t – 16t2at thetime of hitting underneath the distance of basketball
=0 or
2 A) At what time t will the basketball strike the base? B) for just what time t will be the basketball more than 128 base above
of96 base per second. The space s (in base) of basketball from
=> 16t2 – 96 = 0
t
=0 or
=> 16t(t – 6) =0henceeither, 16t
=>t2
=0 or
t
theground after t moments is s(t)= 96t – 16t2
afterthrowing.Answer:- after
=> (t – 4)(t – 2)
fromground = 0 feetso, 96t – 16t2 = 0
8) = 0
equation,
afterthrowing.Answer:- after
will attain underneath.

=>16(t2 – 6t+

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